0$ and $\alpha- \beta\geq2$. It is integrable from the left if and only if $\frac{d_0 (-1)^{\beta-\alpha+1} }{\beta-\alpha+1}<0$ and $\beta-\alpha+1 <0$. Thus, the requirement for left integrability becomes $d_0 (-1)^{\beta-\alpha+1}>0$ and $\alpha- \beta\geq 2$. Hence the weight is both left and right integrable if and only if $d_0>0$, $\beta-\alpha+1<0$, $(-1)^{\beta-\alpha+1}=1$. Thus $\beta -\alpha+1$ is an even negative integer and $d_0 >0$ are the requirements for both integrability from the left and right. Recall that $\frac{b(x)}{a(x)}= d_0 x^{\beta- \alpha} \psi(x)$, with $\psi(0)=1$. Thus $x^{\alpha-\beta} \frac{b(x)}{a(x)}=d_0 \psi(x)$. Hence $ \lim_{x \to0} x^{\alpha-\beta} \frac{b(x)}{a(x)}=d_0$. This gives part (i) of the Proposition - as far as integrability is concerned. Now assume, with the same notation as above and with $r=0$, that $\beta-\alpha+1=0$. In this case, we need to investigate the integrability of $\exp(\int_{p_0}^{x}\frac{2}{n}t^{-1}\psi(t)\,d)/|x^{\alpha}|$. As $\psi(0)=1$, given any $\epsilon>0$, for sufficiently small $x$, $1-\epsilon<\psi(x)<1+\epsilon$. Therefore \[ \frac{2}{n}\,d_0(1-\epsilon)|x|^{-1}<\frac{2}{n}\,d_0 \,|x|^{-1}<\frac{2}{n}\,d_0\,(1+\epsilon)|x|^{-1}. \] Integrating - from the right near $0$, we get \[ K_1\frac{e^{\frac{2}{n}d_0(1-\epsilon)\,ln\,|x|}} {|x^{\alpha}|}<\frac{e^{\int_{p_0}^{x}\frac{2}{n}\,d_0\,t^{-1}\psi(t)dt} }{|x^{\alpha}|}< K_2\frac{e^{\frac{2}{n}d_0(1+\epsilon )\,ln\,|x|}}{|x^{\alpha}|}, \] where $K_1, K_2$ are positive constants. This gives \[ K_1\,|x|^{\frac{2}{n}d_0\,(1-\epsilon)-\alpha}<\frac {e^{\int_{p_0}^{x}( \frac{2}{n}\frac{b(t)}{a(t)}) \,dt} }{|x^{\alpha}|}<2K_2\,|x|^{\frac{2}{n}d_0(1+\epsilon)-\alpha.} \] If the weight function is right integrable near $0$, then necessarily $\frac{2}{n}d_0(1-\epsilon)-\alpha+1>0$. Hence $\frac{2}{n}d_0-\alpha +1>0$. If this holds then the displayed inequalities above establish the integrability of the weight function. Similar arguments give the same condition for integrability from the left- namely $\frac{2}{n}d_0 -\alpha+1 >0$. This completes the proof of the proposition, except for the differentiability of the weight, which is discussed after Proposition \ref{prop3.3} \end{proof} Using lower and upper bounds on the asymptotic form of the weight function (as $x\to\infty$), or partial fraction decomposition of $b/a$, we have the following result. \begin{proposition} \label{prop3.3} (i) If $a$ has no real roots and $p(x)=\exp(\frac{2}{n}\int_{p_0}^{x} \frac{b(t)}{a(t)}dt)/|a(x)|$ then the weight function $p(x)$ gives finite norm for all polynomials if and only if $\deg b-\deg a$ is an odd positive integer and the leading term of $b$ is negative. (ii) If $a$ has only one root, say $0$, then $p(x)=\exp(\frac{2}{n}\int_{p_0}^{x} \frac{b(t)}{a(t)}dt)/|a(x)|$ gives a finite norm for all polynomials on $(0,\infty)$ if and only if \begin{itemize} \item[(a)] $\deg b-\deg a\geq0$ and the leading term of $b$ is negative. \item[(b)] If $a=x^{\alpha}( A_0+A_1x+\dots)$ and $b=x^{\beta}( B_0+B_1x+\dots)$, where $A_0$ and $B_0$ are nonzero constants, then $\alpha-\beta\geq1$, and $\frac{B_0}{A_0}>0$ for $\alpha-\beta\geq2$ whereas $\frac{2B_0}{nA_0}-\alpha+1>0$ for $\alpha-\beta=1$. \end{itemize} \end{proposition} \begin{proof} Assume that $a(x)$ has no real roots. We may assume that the leading term of $a(x)$ is $1$. Thus \[ a(x)=x^n+a_{n-1}x^{n-1}+\dots,\quad b(x)=kx^{m}+\dots, \] where $n, m$ are the degrees of $a,\ b$. Thus $\frac{b(x)}{a(x)}=kx^{m-n}\psi(x)$, with $\lim_{x\to\infty} \psi(x)=1$. Also $a(x)=x^n\phi(x)$ and $\lim_{x\to\infty}\phi(x)=1$. Hence for sufficiently large positive $x$, there are positive constants $d_1, d_2, c_1, c_2$ with \begin{equation} \frac{1}{|x|^nd_1}\,e^{\int_{p_0}^{x}\frac{2}{n} kc_1t^{m-n}dt}

0$, $e^{\int_{p_0}^{x}\frac{2}{n}kct^{m-n} dt}=A|x|^{\frac{2}{n}kc}$. Therefore $\int_{M}^{\infty}Ax^{N}\frac{x^{\frac{2}{n}kc}}{x^n}dx$ cannot be finite if $N$ is large enough. Thus $m-n\neq-1$. Therefore from \eqref{east} we obtain the estimate \[ \frac{A_1}{|x|^n\,d_1}\,e^{\frac{2}{n}kc_1\frac {x^{m-n+1}}{m-n+1}}

0$, where $k$ is the leading term of $b(x)$:
recall that we have taken the leading term of $a(x)$ to be $1$.
Similarly if we require finiteness of the integrals
$\int_{-\infty}^{-M}x^{N} \,e^{\frac{2}{n}kc\frac{x^{m-n+1}}{m-n+1}}\,dx$,
then the requirements are $\frac{k(-1)^{m-n+1}}{m-n+1}<0$, $-(m-n+1)<0$
Therefore the conditions for
$p(x)=\exp(\int\frac{2}{n}\frac{b(x)}{a(x)}dx)/|x^{\alpha}|$
to be a weight on $(-\infty,\infty)$ are that $k<0$, and
$(m-n+1)$ should be an even positive number. This
completes the proof of (i). Part (ii) follows from the integrability
of the weight on $(0,\infty)$ and Proposition \ref{prop3.2}.
\end{proof}
\subsection*{Differentiability properties of the weight function}
The differentiability properties of the weight function at zeroes of the leading
term $a(x)$ follow from the following observation:
if $d$ is a positive integer and $k$ is a positive number, then
$ \lim_{x \to0^{+}} \frac{e^{-kx^{-d}}}{P(x)}=0$ for all polynomials P(x).
Let $N$ be the order of $0$ in $P(x)$. Then $P(x)$ can be written as
$P(x)=ax^{N} Q(x)$ where
$Q$ is a polynomial with $Q(0)=1$. Thus it suffices to show that
$ \lim_{x \to0^{+}} \frac{e^{-kx^{-d}}}{x^{N}}=0$. This is the
same as $ \lim_{u \to\infty} \frac{u^{N}}{e^{ku^{d}}}=0$, which
is obviously true. Therefore
$ \lim_{x \to0^{+}} \frac {e^{-kx^{-d}}}{P(x)} Q(x)=0$ for polynomials
$P, Q$ with $P\ne0$.
As in the proof of proposition \ref{prop3.2}, the weight function, near the root
$r=0$ can then be written as
$p(x)=\psi(x) \exp(\int\frac{2}{n} \frac{b(x)}{a(x)}dx)/(|a_0| |x^{\alpha}|)$
with $\psi(0)=1$. Thus, if $\alpha, \beta$ are the multiplicities
of the root $0$ in $a, b$ respectively and
$\beta- \alpha+1 \ne0$, we have the estimate
\[
\frac{A_1e^{(k_1\frac{2}{n}d_0)( \frac{x^{\beta-\alpha+1}}
{\beta-\alpha+1}) }}{|x^{\alpha}|}\leq\frac{e^{\int^{x}_{p_0}(
\frac{2}{n}\,\frac{b(x)}{a(x)}) \,dt}}{|x^{\alpha}|} \leq\frac
{A_2e^{(k_2\frac{2}{n}d_0)( \frac{x^{\beta-\alpha+1}}{\beta
-\alpha+1}) }}{|x^{\alpha}|}
\]
where $A_1, A_2, k_1, k_2$ are positive constants. We
discuss the right hand limit, as the left hand limit is treated similarly. As
the weight is integrable, we must have $\beta-\alpha+1<0$, $d_0>0$.
Thus, by the above observation, $ \lim_{x \to0^{+}}p(x)=0$, where
$p(x)$ is the weight function. Moreover, by the same observation,
$ \lim_{x \to0^{+}}p(x) R(x)=0$ for any rational
function. Let $\phi(x)=\int^{x}_{p_0}( \frac{2}{n} \frac
{b(x)}{a(x)} ) dt $.
Therefore $p(x)=e^{\phi(x)}/|a(x)|$.
Then all derivatives of
$p(x)$ are of the form $e^{\phi}(x) R(x)$,
where $R(x)$ is a rational function. Therefore the right-hand limits
at $0$ of all the derivatives of the weight function are $0$.
In case $\beta-\alpha+1 = 0$, the weight can be written near a zero
of the leading term - by change of notation- as
\[
p(x) = \frac{c}{|a|} |x|^{( \frac{2}{n}d_0 -\alpha) }
\frac{e^{\phi(x)}}{1+ \psi(x)},
\]
where $\phi,\ \psi$ are analytic functions near $0$, and
\begin{gather*}
a(x)=(x)^{\alpha}(a_0 + a_1 (x) + \dots), \\
b(x)=(x)^{\beta}(b_0 + b_1 (x) + \dots),
\end{gather*}
where $a_0, b_0$ are not $0$ and $d_0=\frac{b_0}{a_0}$.
Here $\frac{2}{n} d_0-\alpha+1>0$ - by the assumption of integrability of
the weight. This completes our discussion of the differentiability
properties of the weight function near a zero of the leading term.
The differentiability properties of ordinary roots have already been
discussed. We now assume that the multiplicity of a root $r$ of
$a(x)=a_n (x)$ is $\alpha$ and its multiplicity in $b(x)=a_{n-1}(x)$
is $\beta$. For convenience of notation we assume that $r$ is zero.
As above, we have $a_n(x)=x^{\alpha}(A_0 + A_1 x+ \dots)$,
$a_{n-1}(x)=x^{\beta}(B_0 + B_1 x+ \dots)$ with $(\beta-\alpha)=-1$. Thus
near $x=0$, the weight is of the form
$p(x)=\frac{1}{|A_0|}|x|^{(\frac{2}{n}\frac{B_0}{A_0}-\alpha) }
\frac{e^{\phi(x)}}{1+\psi(x)}$
where $\phi$, $\psi$ are analytic near zero and $\psi(0)=0$. When there is no
danger of confusion we will write
$p(x)\sim|x|^{( \frac{2}{n}\frac{B_0}{A_0}-\alpha) }$. Now
$p'=p( \frac{2}{n}\frac{b}{a}-\frac{a'}{a} ) $. Therefore, all
higher derivatives of $p$ are of the form $p \rho$ where $\rho$ is a rational
function and all higher derivatives of $p \rho$ are also multiples of
$p$ by rational functions. For later use we record the asymptotic behavior of $p'$
near a zero of $a_n$.
\[
p'(x)=\frac{1}{|A_0|}|x|^{( \frac{2}{n}\frac{B_0}{A_0
}-\alpha) } \frac{e^{\phi(x)}}{1+\psi(x)} \Big( \frac{2}{n}\frac
{b}{a}-\frac{a'}{a} \Big)
\]
The weight $p$ is integrable near zero if and only if $( \frac{2}
{n}\frac{B_0}{A_0}-\alpha+1) >0$. Moreover $ \lim
_{x\to0}p(x) a_n(x) =0$ if and only if
$\frac{2}{n}\frac{B_0}{A_0} >0$.
By the integrability of the weight
$\frac{2}{n}\frac{B_0}{A_0} >\alpha-1 \geq0$. Thus the boundary condition
$ \lim_{x\to0}p(x) a_n(x)=0$ is a consequence of the integrability of the
weight near zero. Similarly
$p(x)a_{n-1}(x)=\frac{1}{|A_0|}|x|^{( \frac{2}{n}\frac{B_0}{A_0
}-1) } \frac{e^{\phi(x)}}{1+\psi(x)} (B_0 + B_1 x + \dots)$,
keeping in mind that $\alpha-\beta= 1$. Hence
$ \lim_{x\to0}p(x) a_{n-1}(x)=0$ if and only if
$( \frac{2}{n}\frac{B_0}{A_0}-1)>0$.
\subsection{Higher order operators}
The principal aim of this section is to prove the following result.
\begin{proposition} \label{prop3.4}
Let $L=a_n(x) y^{(n)}+a_{n-1}(x)y^{(n)} +\dots+ a_2(x) y''+a_1(x) y'$ be a
self-adjoint operator of order $n$ with $n>2$. If $a_n$ has a real root then
the multiplicity of the root is at least $2$ and the multiplicity of the same
root in $a_{n-1}$ is positive and less than its multiplicity in $a_n$.
\end{proposition}
\begin{proof}
Let $r$ be a real root of $a_n$ and assume that it
is a simple root. It is then a logarithmic root. Therefore, near $r$, we have
\[
p(x)\sim|x-r|^{( \frac{2}{n}\frac{B_0}{A_0}-1) },
\]
where $a_n(x)=(x-r) (A_0 + A_1 (x-r)+ \dots)$,
$a_{n-1}(x)= (B_0 +B_1 (x-r)+ \dots)$ and $A_0$, $B_0$ are not zero. Now
$p'=p\big( \frac{2}{n}\frac{a_{n-1}}{a_n}-\frac{a_n'}{a_n} \big) $. Therefore
\[
p'(x)\sim|x-r|^{( \frac{2}{n}\frac{B_{0}}{A_{0}}-1)}
\Big( \frac{2}{n}\frac{a_{n-1}(x)}{a_{n}(x)}-\frac{a_{n}'(x)}
{a_{n}(x)}\Big).
\]
Similarly
\[
p(x) a_{n-1}(x)\sim|x-r|^{( \frac{2}{n}\frac{B_0}{A_0}-1) }
(B_0 + B_1 (x-r)+ \dots)
\]
The boundary conditions and the determining equations in Proposition \ref{prop2.1}
imply that $(a_n p)$, $(a_{n-1} p)$ and $(a_{n-1}p)'$ vanish on the
boundary.
Now $ \lim_{x\to r}a_n(x)p(x)=0$ is a
consequence of the integrability of the weight near $r$. Similarly
$ \lim_{x\to r}a_{n-1}(x)p(x)=0$ if and only if
$( \frac{2}{n}\frac{B_0}{A_0}-1) >0$. Let
$ l_{r} =\lim_{x\to r} (x-r)^{\alpha-\beta} \frac{a_{n-1}(x)}{a_n(x)}$.
Clearly $l_{r}=\frac{B_0}{A_0}=\frac{a_{n-1}(r)}{a_n'(r)}$, as
$\alpha-\beta=1$. Since $ \lim_{x\to r}pa_{n-1}=0$ and
$a_{n-1}(r)\ne0$ we see that $p$ must vanish at $r$ in the sense that its
limit at $r$ is zero. The boundary condition
$ \lim_{x\to r}(a_{n-1}p)'=0$ now implies that
$ \lim_{x\to r}p'(x)=0$. Now
$p'=p( \frac{2}{n}\frac{a_{n-1}}{a_n}-\frac{a'_n}{a_n}) $.
Thus near the root $r$,
\[
p'\sim|x-r|^{(\frac{2}{n}l_{r}-2\alpha)}
\Big( \frac{2}{n}a_{n-1}-a_{n}'\Big).
\]
If $a_{n-1}-\frac{n}{2}a'_n\equiv0$ then in particular
$( \frac{2}{n}\frac{a_{n-1}(r)}{a'_n(r)}-1)=0$.
This means that $ \lim_{x\to r}(x-r)\frac{2}{n}
\frac{a_{n-1}(x)}{a_n(x)}-1=0$ i.e.
$( \frac{2}{n}\frac{B_0}{A_0}-1) =0$. This contradicts the boundary
condition $ \lim_{x\to r}a_{n-1}(x)p(x)=0$.
Let $(a_{n-1} - \frac{n}{2}a'_n)=(x-r)^{\lambda} H(x)$ where
$\lambda\geq0$ and $H(r)\ne0$. If $\lambda> 0$ then
$ \lim_{x\to r}(x-r)\frac{2}{n}\frac{a_{n-1}(x)}{a_n(x)}-1=0$
which again contradicts the boundary condition $ \lim_{x\to r}a_{n-1}(x)p(x)=0$.
Hence $p'\sim|x-r|^{(\frac{2}{n}l_{r}-2\alpha)}H(x)$
so $p'\to0$ at $r$ if and only if $( \frac{2}{n}l_{r}-2\alpha) >0$.
By Proposition \ref{prop2.1}, the operator must satisfy - beside other equations - the
determining equations
\begin{gather}
\label{det1-gen} n(a_np)'=2(a_{n-1}p), \\
\label{det2-gen}\frac{( n-1) (n-2)}{6}(a_{n-1}p)^{\prime\prime
}-( n-2) (a_{n-2}p)'+2(a_{n-3}p)=0
\end{gather}
Equation \eqref{det2-gen} is equivalent to
\begin{equation} \label{rational-gen}
\begin{aligned}
&C_1( \frac{a_{n-1}}{a_n}) ^3
+C_2( \frac{a_{n-1}}{a_n}) ( \frac{a_{n-1}}{a_n}) '
+ C_3( \frac{a_{n-1}}{a_n}) ''\\
&+C_4 ( \frac{a_{n-2}}{a_n}) '+ C_5
\frac{a_{n-1}}{a_n}\frac{a_{n-2}}{a_n}+ C_{6}\frac{a_{n-3}}{a_n}=0
\end{aligned}
\end{equation}
where
\begin{gather*}
C_1=\frac{2(n-1)(n-2)}{3n^2},\quad C_2=\frac{(n-1)(n-2)}{n}, \quad
C_3=\frac{(n-1)(n-2)}{6},\\
C_4= -(n-2),\quad C_5=-\frac{2(n-2)}{n}, \quad
C_{6}=2.
\end{gather*}
This implies the identity
\begin{equation}\label{divide-general}
a_{n-1}\big( a_{n-1}-na_n'\big)
\big(a_{n-1}-\frac{n}{2}a'_n\big) \equiv0 \quad \mod{a_n}
\end{equation}
Using this identity, as $(x-r)$ divides $a_n$ but it does not divide
$a_{n-1}$ nor $( a_{n-1}-\frac{n}{2}a'_n) $, it must
divide $(a_{n-1}-na_n')$. But then
$ \lim_{x\to r}(a_{n-1}-na_n')
=0$. This means that $ \lim_{x\to r}\frac{2}{n}\frac{a_{n-1}
}{a'_n}-2=0$ i.e. $\frac{2}{n}\frac{B_0}{A_0}-2=0$. As seen
above $p'\to0$ at $r$ if and only if $( \frac{2}{n}l_{r}
-2\alpha) >0$. Since $\alpha=1$ we have a contradiction.
Therefore $a_n$ cannot have a simple root and its multiplicity $\alpha$ in
$a_n$ is at least $2$. Suppose that the multiplicity $\beta$ of $r$ in
$a_{n-1}$ is zero. By considering the order of poles of $a_n$ in
\eqref{rational-gen} we see that $\beta$ cannot be zero. This completes the
proof.
\end{proof}
This result has an important consequence for fourth order self-adjoint
operators.
\begin{corollary} \label{coro3.5}
Let $L$ be a self-adjoint operator of order $4$ and $a_4$ be its leading term.
If $a_4$ has more than one real root then it has exactly two real roots
with multiplicity $2$.
Moreover the multiplicity of each real root of $a_4$ in $a_3$ is $1$.
\end{corollary}
\begin{proposition} \label{prop3.6}
Let $n>2$ and suppose that $a=a_n$ has at most one real root.
Then $2\deg b-\deg a\leq n-2$ or $3\deg b-2\deg a\leq n-3$, where $b=a_{n-1}$.
\begin{itemize}
\item[(i)] If $a$ has no real root then
$\deg a<\deg b\leq n-3$;
\item[(ii)] Suppose $a$ has only one real root
$r$ with multiplicity $\alpha$, let $\beta$ be the
multiplicity of $r$ as a root of $b$, and let
$a=(x-r) ^{\alpha}u$, $b=( x-r) ^{\beta}v$.
Then
\[
2\leq\deg a\leq\deg b\leq n-2\quad \text{and}\quad
1+\deg u\leq\deg v\leq n-3
\]
\end{itemize}
\end{proposition}
\begin{proof}
First, in all cases, $\deg b\geq1$. This is because
if $a$ has no real root then $\deg b$ is odd and if $a$ has (at least) one
real root then this will also be a root for $b$ (by Proposition \ref{prop3.4}). If we
multiply by $a^3$ both sides of \eqref{rational-gen}, then the six terms on the left-hand
side will be polynomials with respective degrees
\begin{gather*}
3\deg b,\quad 2\deg b+\deg a-1,\quad 2\deg a+\deg b-2,\\
2\deg a+\deg a_{n-2}-1,\quad \deg a+\deg b+\deg a_{n-2},\quad
2\deg a+\deg a_{n-3}
\end{gather*}
A comparison of degrees shows that $a_{n-2}$ and $a_{n-3}$ cannot be
both zero and that
\[
2\deg b \leq\deg a+\deg a_{n-2},\quad \text{or}\quad
3\deg b \leq2\deg a+\deg a_{n-3}
\]
Using the fact that $\deg a_{j}\leq j$, we obtain
\[
2\deg b-\deg a\leq n-2\quad\text{or}\quad 3\deg b-2\deg a\leq n-3
\]
If $a$ has no real roots then, by Proposition \ref{prop3.3} (i),
$\deg b-\deg a\geq1$ and hence
\[
\deg a<\deg b\leq n-3
\]
If $a$ has only one real root $r$ with multiplicity $\alpha$, then
$\alpha\geq2$ and $b$ has $r$ as a root with multiplicity $\beta$, where
$1\leq\beta<\alpha$ (by Proposition \ref{prop3.4}). Since $\deg b\geq\deg a$, we obtain
that
\[
2\leq\deg a\leq\deg b\leq n-2
\]
Let $a=( x-r) ^{\alpha}u$, $b=( x-r)^{\beta}v$.
Then $\deg a=\alpha+\deg u$ and $\deg b=\beta+\deg v$, and we
obtain $\deg v-\deg u\geq\alpha-\beta\geq1$. Now $\beta+\deg v\leq n-2$, so
$1+\deg u\leq\deg v\leq n-3$ and thus $\deg u\leq n-4$.
\end{proof}
\section{Examples of higher order operators, their eigenvalues and orthogonal
eigenfunctions}
Let $L$ be an operator of the form $L(y)=\sum_{k=0}^na_{k}(x)y^{(k)}$,
where $\deg a_{k}\leq k$; then the eigenvalues of $L$ are the coefficients of
$x^n$ in $L(x^n)$, $n=0,1,2,\dots$.
\begin{proposition} \label{prop4.1}
Let $L$ be a linear operator that
maps the space $\mathbb{P}_n$ of all polynomials of degree at most $n$ into
itself for all $n\leq N$. If the eigenvalues of $L$ are distinct or if $L$
is a self-adjoint operator then there is an eigenpolynomial of $L$ in every
degree $\leq N$.
\end{proposition}
This means that if two operators leave the
space of polynomials of degree at most $n$ invariant for all $n$ and the
weight function which makes the two operators self-adjoint is the same, then
they have the same eigenfunctions. The eigenvalues in general are not simple.
The proof of the above proposition is left to the reader.
Let $\lambda$ be an eigenvalue of $L$ and $\mathbb{P}_n(\lambda)$ the
corresponding eigenspace in the space $\mathbb{P}_n$ of all polynomials of
degree less than or equal to $n$.
If $n_0$ is the minimal degree in $\mathbb{P}_n(\lambda)$, then there is,
up to a scalar only one polynomial in $\mathbb{P}_n(\lambda)$ of degree
$n_0$. Choose a monic polynomial $Q_1$ in $\mathbb{P}_{n_0}(\lambda)$.
Let $n_1$ be the smallest degree, if any, greater than $n_0$ in
$\mathbb{P}_n(\lambda)$. The codimension of $\mathbb{P}_{n_0}(\lambda)$ in
$\mathbb{P}_n(\lambda)$ is $1$. Therefore, in the orthogonal complement of
$\mathbb{P}_{n_0}(\lambda)$ in $\mathbb{P}_{n_1}(\lambda)$, choosing a
monic polynomial $Q_2$, which will necessarily be of degree $n_1$, the
polynomials $Q_1,\ Q_2$ give an orthogonal basis of $\mathbb{P}_{n_1
}(\lambda)$. Continuing this process, we eventually get an orthogonal basis of
$\mathbb{P}_n(\lambda)$ consisting of monic polynomials.
We illustrate this by an example of a fourth order self-adjoint operator that
has repeated eigenvalues but which is not an iterate of a second order
operator. Consider the operator
\begin{equation}\label{op-special-eg}
L=(1-x^2)^2 y^{(4)}-8x(1-x^2) y'''+ 8 y''-24x y'.
\end{equation}
Its eigenvalues are $\lambda_n = n[ (n-1)(n-2)(n+5)-24]$. The eigenvalue
$\lambda= -24$ is repeated in degrees $n=1$ and $n=3$. The weight
function for which this operator is self-adjoint is $p(x)=1$. The
eigenpolynomials of degree at most $3$ are
\[
y_0(x)=1,\quad y_1(x)=x,\quad y_2(x)=x^2-\frac{1}{3}, \quad y_3(x)=x^3.
\]
This gives the set of orthogonal polynomials
$\{1,x,x^2-\frac{1}{3} ,x^3-\frac{3}{5}x\}$.
Since the weight function is the same as that for the
classical Legendre polynomials, this family up to scalars is the same as the
corresponding classical Legendre polynomials.
We now return to examples of higher order operators. The restrictions on the
parameters appearing in all the examples come from integrability of the weight
and boundary conditions. Before giving examples of higher order operators, it
is instructive to consider the classical case of second order operators in the
frame work of section 3.
\subsection{Self-adjoint operators of order $2$}
Assume $n=2$ and that $a_2 (x)$ has distinct roots, which we may assume to
be $-1$ and $1$. If $\alpha$ is the multiplicity of a root $r$ of $a_2 (x)$
and $\beta$ is its multiplicity in $a_1 (x)$ then the integrability of the
associated weight gives the equation $\alpha=\beta+ 1 + \delta$, with
$\delta\geq0$. As $\alpha=1$, we must have $\beta=0$, $\delta=0$. Thus, only
the logarithmic case can occur.
Let $a_1(x)= cx+d$. The integrability condition at a root $r$ reads
$ \lim_{x\to r}(x-r) a_1(x)/ a_2(x)>0$. As we are
assuming that $a_2 (x)=x^2-1$, the integrability conditions at both the
roots gives $c+d>0$, $-c+d<0$, so $-c**
0, \ b+a<0 \text{ and } b\ne0)\\
a_4(x) = (1-x^2)^2\\
a_3(x) = -2(b+(-2+a)x)(-1+x^2)\\
a_2(x) = \frac{b^3+B+2(-1+a)b^2x-Bx^2+b(-2+a+2x^2-3ax^2
+a^2x^2)}{b}\\
a_1(x)= B + \frac{aB x}{b}
\end{gather*}
with the eigenvalues
\[
\lambda_n = \frac{(a-n+1) n ( -b n^2+a b n+b n-a b+{B}) }
{b}.
\]
\subsection{Self-adjoint operators of order $6$}
Consider the self-adjoint operators
\begin{equation}
\label{main-op-n-6}
L=a_{6}(x) y^{(6)}+a_5(x) y^{(5)}+a_4(x) y^{(4)}
+a_3(x) y'''+ a_2(x) y''+a_1(x) y'
\end{equation}
with an admissible weight $p(x)=\exp(\frac{1}{3}\int\frac{a_5(x)}
{a_{6}(x)}dx)/| a_{6}(x)|$, satisfying the differential
equation
\begin{equation} \label{main-eq-n-6}
L(y)=\lambda y.
\end{equation}
By Proposition \ref{prop2.1}, the determining equations in this case are
\begin{gather}
3(a_{6}p)'= (a_5 p)\label{det1-n-6}\\
5(a_5p)''- 6(a_4p)'+ 3(a_3p)=0 \label{det2-n-6n}\\
(a_4p)'''-3(a_3p)''+5(a_2p)'-
5(a_1p)=0 \label{det3-n-6n}
\end{gather}
on $I$, subject to the vanishing of
\begin{equation}\label{boundary-n-6}
(a_{6} p),\quad(a_5 p), \quad
(a_5p)', \quad(a_4p), \quad(a_4p)'- 3(a_3p), \quad
(a_4p)''-3(a_3p)'+5(a_2p)
\end{equation}
on the boundary $\partial I$. Equations \eqref{det2-n-6n} and
\eqref{det3-n-6n} are equivalent to
\begin{equation}
\frac{10}{27}\big(\frac{a_5}{a_{6}}\big) ^3
+\frac{10}{3}\big(\frac{a_5}{a_{6}}\big) \big( \frac{a_5}{a_{6}}\big) '+
\frac{10}{3}\big( \frac{a_5}{a_{6}}\big) ''
-4 \big(\frac{a_4}{a_{6}}\big) '-\frac{4}{3} \frac{a_5}{a_{6}}
\frac{a_4}{a_{6}}+ 2\frac{a_3}{a_{6}}=0
\end{equation}
and
\begin{align*}
& -81\big( \frac{{a_3} }{{a_{6}} }\big) ''
+\frac{{a_4}}{{a_{6}} } \big( \frac{{a_5} }{{a_{6}} }\big) ^3
-9 \frac{{a_3}}{{a_{6}} } \big( \frac{{a_5} }{{a_{6}} }\big) ^2
+45 \frac{{a_2} }{{a_{6}} } \big( \frac{{a_5} }{{a_{6}} }\big)
-135 \frac{{a_1}}{{a_{6}} }+135 \big( \frac{{a_2} }{{a_{6}} }\big) '\\
&+27 \big( \frac{{a_4} }{{a_{6}} }\big) '''
+9 \frac{{a_4} }{{a_{6}} } \big( \frac{{a_5} }{{a_{6}} }\big)''
+27 \frac{{a_5} }{{a_{6}} } \big( \frac{{a_4} }{{a_{6}}}\big) ''
-54 \big( \frac{{a_5} }{{a_{6}} }\big) \big(\frac{{a_3} }{{a_{6}} }\big) '\\
& +9 \big( \frac{{a_5} }{{a_{6}}}\big) ^2 \big( \frac{{a_4} }{{a_{6}} }\big) '
+9\frac{{a_4} }{{a_{6}} } \frac{{a_5} }{{a_{6}} }
\big( \frac{{a_5} }{{a_{6}} }\big) '
-27 \big( \frac{{a_3} }{{a_{6}} }\big) \big( \frac{{a_5}
}{{a_{6}} }\big) '+27 \big( \frac{{a_4} }{{a_{6}} }\big)'
\big( \frac{{a_5} }{{a_{6}} }\big) '=0.
\end{align*}
These identities give the congruences
\begin{gather} \label{divide-n-6-1}
5 {a_5} ( {a_5} -6 {a_{6}}')
( {a_5} -3 {a_{6}}') \equiv0 \quad\mod{a_{6}}\\
\label{divide-n-6-2}{a_4} ( {a_5} -9 {a_{6}}')
( {a_5} -6 {a_{6}}') ( {a_5} -3 {a_{6}}') \equiv0 \quad \mod{a_{6}}.
\end{gather}
Before presenting examples, we note that, in case the leading term has a real
root, there are many operators that satisfy the determining equations but for
which one of the boundary condition fails at one or both end points of
interval $I$; in case the leading term has no real roots, the boundary
conditions will be satisfied - because of the form of the weight - but
one of the determining equations will not be satisfied. Here are some typical
examples.
\begin{example} \label{examp4.1}\rm
The operator in \eqref{main-op-n-6} with
\begin{gather*}
p(x) = (x-1)^2 (x+1),\\
a_{6} (x) = (x-1)^2 (x+1)^{4}, \quad
a_5 (x) = 3 (x-1) (x+1)^3 (9 x-1),\\
a_4(x) = 60 x (x+1)^2 (5 x-3), \quad
a_3(x) = 240 ( 7 x^3+6 x^2-2 x-1) ,\\
a_2(x) = 720 x (5 x+3), \quad a_1(x) = 360 (x+1)
\end{gather*}
satisfies the determining equations \eqref{det1-n-6}, \eqref{det2-n-6n},
\eqref{det3-n-6n} and all the boundary conditions in \eqref{boundary-n-6}
except that the last boundary condition fails at the end point $1$ of
$I=[-1,1]$. Therefore, these coefficients and weights would not constitute a
self-adjoint operator.
\end{example}
\begin{example} \label{examp4.2} \rm
The operator in \eqref{main-op-n-6} with
\begin{gather*}
p(x) = e^{-m^2 x} x^2 \quad (m\ne0)\\
a_{6} (x) = x^2, \quad a_5 (x) = -3 x (m^2 x-4),\\
a_4(x) = -30 m^2 x+{A} x^2+30, \quad a_3(x) = 5 x^2 m^{6}-2 ({A}
x^2+30 ) m^2+8 {A} x,\\
a_2(x) = x ({C}-3 m^{8} x)+{A} (m^{4} x^2 +12),\\
a_1(x) = 18 m^{8} x-60 m^{6}-8 {A} m^{4} x+24 {A} m^2+{C} (3-m^2 x)
\end{gather*}
satisfies the determining equations \eqref{det1-n-6}, \eqref{det2-n-6n},
\eqref{det3-n-6n} and all the boundary conditions in \eqref{boundary-n-6}
except that the last boundary condition fails at the end point $0$ of $I =
[0,\infty)$. Again these coefficients and weights would not constitute a
self-adjoint operator.
\end{example}
\begin{example} \label{examp4.3} \rm
The operator in \eqref{main-op-n-6} with
\begin{gather*}
p(x) = \frac{e^{-m^2 x^2}}{x^2+1} \quad(m\ne0)\\
a_{6} (x) = x^2+1, \quad a_5 (x) = -6 m^2 x (x^2+1),\\
a_4(x) = 10 x^{4} m^{4}-10 m^{4}+{A} x^2+{A}, \\
a_3(x) = -4 m^2 (-10 m^{4}+5 m^2+{A}) x (x^2+1),\\
a_2(x) = {C_2} x^2+{C_1} x+{C_0}, \quad a_1(x)
= {D_0 }+{D_1} x
\end{gather*}
satisfies the determining equations \eqref{det1-n-6}, \eqref{det2-n-6n} and
all the boundary conditions in \eqref{boundary-n-6} for $I = (-\infty,\infty)$
but fails to satisfy the remaining determining equation \eqref{det3-n-6n}.
\end{example}
Examples of sixth order self-adjoint operators with the weights of the
form $p(x)=e^{-x^2}$, $p(x)=|x|^n e^{m x}$ and $p(x)=\frac{(1+x)^{m}
}{(1-x)^n}$ can be found, similar to fourth order case, by solving the
determining equations and boundary conditions using Mathematica. However,
because of space constraint, the long expressions for operators and
eigenvalues are not reproduced here, and are provided in expanded online
version of the paper at http://arxiv.org/abs/1409.2523.
\subsection{Self-adjoint operators of order $8$}
Consider the self-adjoint operator
\begin{equation} \label{main-op-n-8}
\begin{aligned}
L&=a_{8}(x) y^{(8)}+a_{7}(x) y^{(7)}+a_{6}(x) y^{(6)}
+a_5(x) y^{(5)}+a_4(x) y^{(4)}\\
&\quad +a_3(x) y'''+ a_2(x) y''+a_1(x) y'
\end{aligned}
\end{equation}
with an admissible weight $p(x)=\exp(\frac{1}{4}\int\frac{a_{7}(x)}
{a_{8}(x)}dx)/| a_{8}(x)|$, satisfying the differential
equation
\begin{equation}
\label{main-eq-n-8}L(y)=\lambda y.
\end{equation}
By Proposition \ref{prop2.1}, the determining equations in this case are
\begin{gather}
4(a_{8}p)'= (a_{7} p)\label{det1-n-8}\\
7(a_{7}p)''- 6(a_{6}p)'+ 2(a_5p)=0 \label{det2-n-8n}\\
(a_{6}p)'''-2(a_5p)''+2(a_4p)'-(a_3p)=0\label{det3-n-8n}\\
(a_5p)^{(4)} - 4 (a_4p)'''+ 9(a_3p)''-14(a_2p)'+ 14 (a_1p)=0 \label{det4-n-8n}
\end{gather}
on $I$, subject to the vanishing of
\begin{equation} \label{boundary-n-8}
\begin{gathered}
(a_{8} p),\quad (a_{7} p), \quad (a_{7}p)', \quad
(a_{6}p), \quad (a_{6}p)', \quad (a_5 p), \\
5(a_{6}p)''-11(a_5p)' +14(a_4p),\quad 9(a_{6}p)''-17(a_5p)'+14(a_4p), \\
(a_5p)''-4(a_4p)'+9(a_3p), \quad
(a_5p)'''- 4(a_4p)''+9(a_3p)'-14(a_2p)
\end{gathered}
\end{equation}
on the boundary $\partial I$.
The examples of eighth order self-adjoint operators with the weights of the
form $p(x)=e^{-x^2}$, $p(x)=|x|^n e^{m x}$ and
$p(x)=\frac{(1+x)^{m}}{(1-x)^n}$ can be found, similar to fourth order case,
by solving the determining equations and boundary conditions using Mathematica.
Because of space constraint, the long expressions for operators and eigenvalues
are not reproduced here, and instead are provided in expanded online version of the
paper at http://arxiv.org/abs/1409.2523.
\subsection*{Acknowledgements}
We thank Professor Mourad Ismail for his valuable comments and suggestions
during the preparation of this paper.
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\end{document}
**